Look..It's easy..
du/dx=f(x)..(this is where you are making mistake).The correct equation is du/dx= -f(x)
so f(x)= -2cx(This force is opposite to the direction of velocity of particle.i.e.the force is in negative x-direction..
Now do the process again and you will easily reach the answer..

If you insist on using Newton's second law, then I should say that this step of yours: ## \int 2cxdt = \int m dv \Rightarrow \frac {2cxt} m = v + \text{constant} ## is already wrong, and not really because of the missing minus sign, but because ##x## is a function of time, no a constant, so its integral is not ##xt##.

If you have studied linear differential equations, you could also observe that Newton's second law gives $$ \frac {d^2 x} {dt^2} + \frac {2 c} {m} x = 0 $$ which is a second-degree linear diff. eq., whose general solution is $$ x = A \sin (\sqrt{\frac {2c} m}t + \alpha) $$ which is again periodical with the same period, of course.

I greatly appreciate your help. Your solution is a work of art. I will plug in some hypothetical numbers to make sure our solutions are equivalent (unless you have already checked this). I need to write a solution in my own way and I simply would not legitimately be able to reproduce your solution on my own. Although I have saved this solution and will use it to improve with time.